Question

Two particles A and B are projected with same speed so that ratio of their maximum heights reached is $3:1$. If the speed of A is doubled without altering other parameters, the ratio of horizontal ranges attained by A and B is?

Answer 1

We know maximum height of a projectile motion is given by $H=\frac{{v_0}^2{\sin }^2\theta }{2g}$

Given ratio of maximum height attained by A and B=>$\frac{H_A}{H_B}=\frac{\frac{{v_{0A}}^2{\sin }^2{\theta }_A}{2g}}{\frac{{v_{0B}}^2{\sin }^2{\theta }_B}{2g}}=\frac{3}{1}$

As given velocity of A and B are same thus the above equation reduces to

$\frac{{\sin }^2{\theta }_A}{{\sin }^2{\theta }_B}=\frac{3}{1}\frac{\sin {\theta }_A}{\sin {\theta }_B}=\frac{\sqrt{3}}{1}$

We can conclude fro the above equation that the values of ${\theta }_{A}and{\theta }_B$ are ${60}^{0}and{30}^0$ respectively.

Now the ratio of horizontal range covered by A and B

$\frac{R_A}{R_B}=\frac{\frac{({V)}_{0A}^2\sin 2\theta }{g}}{\frac{({V)}_{0B}^2\sin 2\theta }{g}}$

Now as per question the velocity of A is doubled,thus the above equation reduces to

$\frac{R_A}{R_B}=\frac{\frac{{4V}_{0A}^2\sin 2\theta }{g}}{\frac{V_{0A}^2\sin 2\theta }{g}}=\frac{4\sin 120}{\sin 60}=4\frac{0.866}{0.866}=4m$